Confirm this

[alkatran]

I was bored at work (as usual ) and tried to do a math problem. I had to figure out what the function was.

f(1) = 23
f(2) = (other number, im just making these up)
f(4) = ...

Anyways, I solved it by changing it to

a(1^2) + b(1) + c = 23
a(2^2) + b(2) + c = ...

etc.
Anyways, I thought about it afterwards, and was jsut wondering if I need a max of an xth degree polynomial (assuming that's spelled right, french math here) where x is the number of points given (there were 3 in this example).

(need a constant for one point, a first degree for 2 (mx + a), etc)

[jemidiah]

My own ramblings :

Take three points in a line, and now try passing a parabola through them; you'd need a line. In that case, it would be less than or equal to n-1. Hmm, maybe given any n points of a function, there is (at least?) one n-1 or less degree polynomial that can have those points in its solution set?

As another test, let's try taking 5 points where the first three go (relatively) up and the last 2 go (relatively) down. Well, I doubt a parabola would work except in special cases, though there are so many possibilities with cubics, quartics, and quintics that I'd bet one of them would work.

Anyhow, it'd be an interesting premise to research more.

[alkatran]

I've convinced myself that it's true, and applied it to 2 points and 3 points. (btw, I did say that you needed a MAX of xth degree, so 3 points could be a line).

You definitely can't use a quadratic to solve 4 points, if they are above, below, above, and below 0 (3 cuts across y=0).

[RAEsquivelC]

I believe jemidiah is correct in stating that you need a polynomial of a maximum degree of n-1 to assure all n points lie on it. for example:
For 2 points, a straight line will do, degree = n-1 = 1
For 3 points, a circle, parbola, an ellipse, any 2nd degree curve will be sufficient, or, n-1 = 2
For 4 points, a third-degree polynomial will sufice.
Etc.

[sql_lall]

Lagrange Interpolation

mathworld it...u'll be pleasantly surprised!

[alkatran]

Originally posted by sql_lall
Lagrange Interpolation

mathworld it...u'll be pleasantly surprised!

Pleasantly surprised that I was right?

How old is this thread, anyways?

[sql_lall]

Not sure how old this was, but the previous post before mine was recent....i was just answering that one.

And yeah, i just think that Lagrange Interpolation is one of the most elegant, simple, and logical formula ever. It's so neat (except when having to cancel down), and so obvious once you see it. I've come to the conclusion that Lagrange was a genius.